Unit - 7
Mechanical Vibrations [10ME72]
Vibration of Continuous System
UNIT
UNIT - 7- 8
Dr. T. Jagadish. Professor
MODAL forANALYSIS
Post Graduation, Department
AND CONDITION of Mechanical Engineering, Bangalore
MONITORING
MULTI DEGREE
MULTIDEGREE FREEDOM
VIBRATIONS OFSYSTEMS
SYSTEMS
Institute of Technology, Bangalore
Continuous systems are tore which have continuously distributed mass and Elasticity. These continuous
systems are assumed to be homogeneous and isotropic obeying hook law within the elastic limit. Since to
specify the position of every point in the continuous systems an infinite number of coordinates is required
therefore a continuous systems is considered to have infinite number of degrees of freedom. Thus there will
be infinite natural frequencies
Free vibration of continuous system is sum of the principal or normal modes. For the normal mode
vibration, every particle of the body performs simple harmonic motion at the frequency corresponding to
the particular root of the frequency equation, each particles passing simultaneously through its respective
equilibrium position. If the elastic curve resulting due to vibration motion starts coinciding exactly with
one of the normal modes, only that normal mode will be produced. If the elastic curve of the system under
which the vibration is started, is identical to any one of the principal mode shapes, then the system will
vibrate only in that principal mode.
Vibration of String.
A string stretched between two support is shown in figure-1 is an infinite degree of freedom system.
T
Y θ + (δθ
(δθ/δx)
/δ dx
y
θ dx
x dx X
T
Figure-1
Dept. Of ME, ACE 1
Mechanical Vibrations [10ME72]
Let the tension T in the string be large so that for small displacement or amplitude the tension in the
string remains constant throughout the string. Consider an element of length dx at a distance ‘x’ from the
left end. At any instant of time the element of the string be displaced through a distance ‘y’ from the
equilibrium position. Then the tension at both the ends of this element is ‘T’. If θ the angle at the left end
of the element makes with the horizontal x- axis then the angle at the right end of the element is
θ + (δθ/δx)dx
The components of these two tensions at the ends of the element along x-axis balance each others.
The components along Y – axis
sin Since θ is small
1
If ‘ρ' is mass per unit length, then the mass of the element is ‘ρdx’ and the differential equation of motion
along Y-axis according to Newton’s second law of motion is,
Mass x Acceleration = Sum of all the forces.
Therefore
But we know slope then we have
or or
⁄
--- (2)
Where !⁄ ------ (3) is the velocity of wave propagation along the string. Solution of equation (2)
can be obtained by assuming Y through separation of variables to be a product of two functions in the
form. " , $ % & $ 4
()" ()" ()" ()"
& * % 5
( ) ( ) ($ ) ($ )
Substituting equation (5) in to equation (2) we have
2
Dept. Of ME, ACE 2
Mechanical Vibrations [10ME72]
, - - 0 -
& ./ ---- (6)
,
- ,
-
Left hand side of equation (6) is a function of x alone and the right hand side is a function of t alone. These
two can only be equal if each one of these expression is a constant. This constant may be positive, zero or
negative. If it is a positive constant or zero then there is no vibratory motion. Hence this constant has to be
2
negative constant and equal to -
0 -
Then ) 7
,
-
)
()" 1 ()&
2) 0 * 2) 0
% ( ) & ($ )
()" 4 ()&
) % 0 * 2) 0 8
( ) ($ )
With the general solution Y A sin 8
9
: ;.< 89 9
G D sin$ @;.<$ 10
Substituting equation (9) and (10) into equation (4) we have
4 4
" , $ A sin 0
: ;.< 0
D sin$ @;.<$ 11
Since 2 can have infinite values, being an infinite degree of freedom system the general solution can be
written as
" , $ ∑∞
EF BAisin 9 : ;.< 9 DDisin$ @;.<$ 11
8C 8C
The arbitrary constant A, B, D and E have to be determined from the boundary condition and the initial
conditions.
For the string of length ‘L’ stretched between two fixed points. The boundary conditions are
y(0,t) = y(L, t) = 0
The condition y (0, t) = 0 will require B = 0 so the solution will be
Dept. Of ME, ACE 3
Mechanical Vibrations [10ME72]
4
" , $ A sin 0
D sin$ @;.<$ 12
The condition y (L, t) = 0 them fields the equation 0 BA sin H DD sin$ @;.<$
8
9
Since I<2$ @;.<J$ K 0, then we have BA sin 8
9
HD0
or B89H D L for n = 1, 2,3 ………. ----- (13)
Since M ;/O ------------ (14) and
2LM 2L;/P ---------- (15). Where λ is the wave length of oscillation.
Substituting equation (15) into equation (13) we have
BQR/P
9
DH L or 2H/O -------- (16)
Each ‘n’ represents normal mode vibration with natural frequency determined from the equation
S
2HM/; therefore M ;/2H = )T √/V for n = 1,2,3…….. ------ (17)al with the
distribution % B sin L
H
D ----- (18)
Thus the general solution is given by
∞
S\
" , $ WI sin t @ cos t sin T
19
1
S\
With 20
T
With the initial condition of Y(x,0) and Y (x, 0) the constants Dn and En can be evaluated.
Longitudinal vibration of Rod/Bar
Dept. Of ME, ACE 4
Mechanical Vibrations [10ME72]
The bar/rod considered for the analysis is assumed to be thin and a uniform cross sectional area throughout
its length. Due to axial force the displacement ‘u’ along the bar will be a function of both position ‘x’ and
time ‘t’. The bar/rod has infinite number of natural frequencies and modes of vibration
Consider an element of length dx at a distance ‘x’ from the left end of the bar as shown in the
figure -2.
igure -2 Displacement in rod/bar member.
Figure
At any instant of time during vibration let ‘P’ be the axial force at the left end of the element. Then the
axial force at the right end of the element is P+(
P+(∂p/∂x)dx
x)dx and If ‘u’ is the displacement at x, then the
displacement at x + dx will be u+(∂u/
u/∂x)dx.
x)dx. Then the element dx in the new position has changed in length
x)dx and hence the unit strain is ∂u/∂x.
by an amount of (∂u/∂x)dx
But from Hook’s law the ratio of unit stress to the unit strain is equal to the Modulus of Elasticity ‘E’,
‘E
Hence we have ∂u/∂xx = P/AE ------- (1)
In which ‘A’ is the cross sectional area of the bar/rod. Differentiating the equation (1) with
w respect to x we
have AE ∂2u/∂x2 = ∂P/∂x ------ (2) But (∂P/∂x) dx is the unbalanced force
Considering the Dynamic Equilibrium of the element from Newton’s second law of motions me have.
(Mass) X (Acceleration) of the Element = (Unbalanced Resultant External force)
(∂P/∂x) dx = Ar dx ∂2u/∂t2 = ------ (3)
ass per unit volume, substituting equation (2) into (3) for ∂P/∂x we have
Where ρ is the density of rod/bar mass
EAdx (∂2u/∂x2) = rAdx ∂2u/∂t2
∂2u/∂x2 = (r/E) ∂2u/∂t2
5
Dept. Of ME, ACE
Mechanical Vibrations [10ME72]
∂2u/∂x2 = [1/(E/r)] ∂2u/∂t2 --------------- (4)
∂2u/∂x2 = (1/c2) ∂2u/∂t2 --------------- (5)
Where c = √(E/r) ------ (6) is the velocity of propagation of the displacement or stress wave in the rod/bar.
Solution of equation (5) can be obtained by assuming u through separation of variables to be a product of
two functions in the form u(x, t) = U(x) G(t) ------ (7)
∂2u/∂x2=G(∂2U/∂x2) and ∂2u/∂t2=U(∂2G/∂t2) --(8)
Substituting equation (8) in to equation (5) we have G(∂2U/∂x2) = (U/c2)(∂2G/∂t2)
Then (c2/U)(∂2U/∂x2) = (1/G)(∂2G/∂t2) ----- (9)
Left hand side of equation (9) is a function of x alone and the right hand side is a function of t alone. These
two can only be equal if each one of these expression is a constant.
This constant may be positive, zero or negative. If it is a positive constant or zero then there is no vibratory
motion. Hence this constant have to be negative constant and equal to - w2 . Then we have
(c2/U)(∂2U/∂x2)=(1/G)(∂2G/∂t2)= - ω2 ------- (10)
(c2/U)(∂2U/∂x2)= - w2 and (1/G)(∂2G/∂t2)= - ω2
∂2U/∂x2+(ω/c)2U = 0 and ∂2G/∂t2+ω2G = 0 - (11)
With the general solution U = Asin(ω/c)x + Bcos(ω/c)x ----- (12) G = Dsinωt + Ecosωt ------ (13)
Substituting equation (12) and (13) into equation (7) we have
u(x, t) = [Asin(ω/c)x + Bcos(ω/c)x] * [Dsinωt + Ecosωt] --------- (14)
Since ω can have infinite values, as the system being an infinite degree of freedom the general solution can
be written as
u(x, t) =∑i∞=1[Aisin(ωi/c)x + Bicos(ωi/c)x] [Disinωit + Eicosωit ] -------- (15)
The arbitrary constant A, B, D and E have to be determined from the boundary condition and the initial
conditions.
Dept. Of ME, ACE 6
Mechanical Vibrations [10ME72]
Example:
Determine the natural frequencies and mode shape of a bar/rod when both the ends are free.
free
Solution: When the bar/rod with both the ends being free then the stress and strain at the ends are zero
The boundary conditions are ∂u/∂x
∂x = 0 at x= 0 and x= L ------- (1)
u(x, t) = [Asin(ω/c)x + Bcos(ω/c)x]
/c)x] [Dsin
[Dsinωt + Ecosωt] ------------ (2)
∂u/∂x = [(Aω/c)cos(ω/c)x + (Bω/c)sin(
/c)sin(ω/c)x] [Dsinωt + Ecosωt] ------------ (3)
(∂u/∂x)x=0=(Aω/c)[Dsinwt + Ecosωt]
t] = 0 --- (4)
(∂u/∂x) x=L = (ω/c)[Acos(ωL/c)
L/c) + Bsin(
Bsin(ωL/c)] [Dsinωt + Ecosωt] = 0 ------ (5)
Since the equation (4) and (5) must be true for any time ‘t’
From equation (4) A must be equal to zero. Since B must be finite value in order to have vibration then
sin(ωnL/c) = 0 or ωnL/c =sin-1(0) = nπ
equation (5) is satisfied only when sin(
Since c = √(E/ρ) then we have ωnL
L√(ρ/E) = nπ Thus ωn = (nπ/L) √(E/ρ) ------ (6)
since the natural frequency is fn = ωn/2π = [(nπ/L) √(E/ρ)]/ 2π
fn = (n/2L) √(E/ρ) ------------- (7)
Each ‘n’ represents normal mode vibration with natural frequency ‘fn’ determined from the
equation (7).
Torsional
sional vibration of circular shaft.
Consider an element of length dx at a distance x from one end of the shaft as shown in the figure -3
7
Dept. Of ME, ACE 7
Mechanical Vibrations[10ME72]
Figure -3 Torque acting on the element of the circular shaft
At any given instant of time during vibration let T be the torque at left end of the element. Then the torque
at the right end of the element is given by T+(∂T/∂x)dx.
If θ is the angular twist of the shaft at the distance x, then θ+(∂θ/∂x)dx is the angular twist of the
shaft at the distance x+dx. Therefore the angular twist in the element of length dx is (∂θ/∂x)dx.
If ‘J’ is the polar moment of inertia of the shaft and ‘G’ the modulus of rigidity then the angular twist in the
element of length dx is given by torsion formula
dθ = (T/GJ) dx ------- (1) or GJ(dθ/dx) = T ------- (2)
Differentiating with respect to x we have GJ(d2θ/dx2) = dT/dx ------- (3)
In which GJ is the torsional stiffness of the shaft. Since the torque on the two face of the element being T
and T+(∂T/∂x)dx the net torque on the element will be (∂T/∂x)dx --- (4)
Substituting equation (3) for ∂T/∂x in to equation (4) we have (∂T/∂x)dx = GJ(∂2θ/∂x2)dx ------- (5)
Considering the dynamic equilibrium of the element one can obtain the equation of motion by equating the
product of mass moment of inertia ‘Jρdx’ and the acceleration ‘∂2θ/∂t2’ to the net torque acting on the
element we have.
(Jρdx) ∂2θ/∂t2 = GJ(∂2θ/∂x2)dx
∂2θ/∂t2 = (G/ρ)(∂2θ/∂x2) or
∂2θ/∂x2 = (ρ/G) ∂2θ/∂t2
∂2θ/∂x2 = [1/(G/ρ)] ∂2θ/∂t2
∂2θ/∂x2 = (1/c2) ∂2θ/∂t2 --------- (6)
Where c = √(G/ρ) ------ (7) is the velocity of wave propagation in which r is the density of the shaft in
mass per unit volume.
Dept. Of ME, ACE 8
Mechanical Vibrations[10ME72]
Solution of equation (6) can be obtained by assuming q through separation of variables to be a product of
two functions in the form. θ(x, t) = Θ(x) G(t) ------ (7)
∂2θ/∂x2=G(∂2Θ/∂x2) and ∂2θ/∂t2=Θ(∂2G/∂t2)---(8)
Substituting equation (8) in to equation (6) we have G(∂2Θ/∂x2) = (Θ/c2)(∂2G/∂t2)
Then (c2/Θ)(∂2Θ/∂x2) = (1/G)(∂2G/∂t2) ----- (9)
Left hand side of equation (9) is a function of ‘Θ’ alone and the right hand side is a function of ‘t’ alone.
These two can only be equal if each one of these expression is a constant. This constant may be positive,
zero or negative. If it is a positive constant or zero then there is no vibratory motion. Hence this constant
has to be negative constant and equal to - ω2
(c2/Θ)(∂2Θ/∂x2)=(1/G)(∂2G/∂t2) = -ω2 -- (10)
(c2/Θ)(∂2Θ/∂x2)= - ω2 and (1/G)(∂2G/∂t2)= - ω2
∂2Θ/∂x2+(ω/c)2Θ = 0 and ∂2G/∂t2+ω2G = 0--(11)
With the general solution Θ = Asin(ω/c)x + Bcos(ω/c)x ----- (12) G = Dsinωt + Ecosωt ------ (13)
Substituting equation (12) and (13) into equation (7) we have
θ(x, t) = [Asin(ω/c)x + Bcos(ω/c)x] [Dsinωt + Ecosωt] --------- (14)
Since ω can have infinite values, as the system being an infinite degree of freedom the general solution can
be written as
θ(x, t) =∑i∞=1[Aisin(ωi/c)x + Bicos(ωi/c)x] [Disinωit + Eicosωit ] -------- (15)
The arbitrary constant A, B, D and E have to be determined from the boundary condition and the initial
conditions.
Lateral Vibration of beams
To derive the differential equation of motion for lateral vibration of beams one has to consider the forces
and moments acting on the beam. Let V and M are the shear forces and bending moments respectively
acting on the beam with p(x) represents the intensity of lateral loading per unit length of the beam.
Dept. Of ME, ACE 9
Mechanical Vibrations[10ME72]
Consider a beam of length ‘L’ and moment of inerter ‘I’ subjected to distributed lateral load of p(x) N per
unit length. At any section x-x at a distance x from one end consider an element of length dx as shown in
the figure-4. The force and the moments acting on the element of length dx are as shown in the figure-5
Figure-4 Beam Figure-5 Force and Moments
For static equilibrium the summation of the forces and moment should be equal to zero. Thus summing the
vertical force in the Y – direction and equating to zero we have
V + p(x) dx – [V +(∂V/∂x) dx] = 0
[p(x) - (∂V/∂x)] dx = 0 since dx ≠ 0
p(x) - (∂V/∂x) = 0 or ∂V/∂x = p(x) ---- (1)
Which state that the rate of change of shear force along the length of the beam is equal to the load per unit
length.
Summation of the moment about any point should be equal to zero. Thus summing the moments about the
right bottom corner of the element considering the clockwise moment as positive we have
M + p(x) dx2/2 + V dx – [M +(∂M/∂x) dx] = 0 or p(x) dx2/2 + V dx – (∂M/∂x) dx = 0
Since dx is small neglecting the higher order terms we have
[V – (∂M/∂x)] dx = 0 Since dx ≠ 0 we have
V – (∂M/∂x) = 0 ∂M/∂x = V ----- (2)
Which state that the rate of change of bending moment along the length of the beam is equal to the shear
force.
From equation (1) and (2) we have ∂2M/∂x2 = ∂V/∂x = p(x) ----- (3)
Dept.. Of ME, ACE 10
Mech
Mechanical Vibrations
V ibrations [10Me72]
[10ME72]
But we known that the bending moment is related to curvature by flexure formula and for the coordinates
indicated we have EI(d2y/dx2) = M ----- (4)
Substituting equation (4) into equation (3) we have ∂2[EI(d2y/dx2) ]/∂x2 = p(x)
EI (d4y/dx4) = p(x) ----- (5)
For the dynamic equilibrium when the beam is having transverse vibrations about its static equilibrium
position under its own weight and due to the distributed load per unit length must be equal to the inertia
load due its mass and acceleration. By assuming harmonic motion the inertia force ρωy2 is in the same
direction as that of the distributed load p(x).
Thus we have p(x) = ρω2y ------ (6)
In which ρ is the mass density per unit length of the beam. Thus substituting equation (6) into equation (5)
for p(x) we have EI(d4y/dx4) = ρω2y or EI(d4y/dx4) - ρω2y = 0 ------ (7)
If the flexural rigidity EI is constant then equation (7) can be written as d4y/dx4 - (ρω2/EI)y = 0 ---- (8)
Letting β4 = ρω2/EI ------ (9) d4y/dx4 – β4y = 0 ------ (10)
The above equation (10) is a fourth order differential equation for the lateral vibration of a uniform cross
section beam. The solution of the above equation (10) can be obtained by assuming the displacement ‘y’ of
the form y = eax ---- (11)
Which will satisfy the fourth order differential equation (10) when a = ± β and a = ± iβ ---- (12)
Since e±βx = coshβx ± sinhβx ---- (13) and e±ibx = cosβx ± sinβx ---- (14)
Then the solution of the above equation (10) will be in the form
Y=Acoshβx+Bsinhβx+Ccosβx+Dsinβx -- (15) Which is readily established
The natural frequencies of lateral vibration of beams are found from equation (9) given by
βn4 = ρωn2/EI Thus ωn2 = βn4 EI/ρ or ωn = βn2√EI/ρ
ωn = (βnL)2√(EI/ρL4) ------ (16) Where the number βn depends on the boundary conditions of the beam.
The following table lists the numerical values of (βnL)2 for typical end boundary conditions of the beam.
Dept. Of ME , ACE 11
Mechanic al Vibrations [10ME72]
Beam Condition First Mode Second Mode Third Mode
Simply supported 9.87 39.5 88.9
Cantilever 3.52 22.0 61.7
Free-Free 22.4 61.7 121.0
Clemped -Clemped 22.4 61.7 121.0
Champed hinged 15.4 50.0 104.0
Hinged -Fre 0 15.4 50.0
Example-1: Determine the first three natural frequencies of a Rectangular Cantilever Beam for the
following data Length L = 1 m Breath B = 0.03 m Depth D = 0.04 m Young's Modulus
E = 2x1011 N/m2 Density r = 7850.0 Kg/m3
Solution: For a cantilever beam the boundary conditions are at x= 0 displacement and slope are zero
i.e. y = 0 and dy/dx = 0 ------ (1) and at x= L the shear force ‘V’ and the bending moment ‘M’ is zero
i.e. d2y/dx2 = 0 and d3y/dx3 = 0 -------- (2)
Substituting these boundary conditions into the general solution
Y=Acoshβx + Bsinhβx + Ccosβx + Dsinβx --- (3)
we obtain (y)x=0 = A + C = 0 then A = - C – (4)
(dy/dx)x=0= β(Asinhβx + Bcoshβx – Csinβx + Dcosβx)x=0 = 0
β(B + D) = 0 since β ≠ 0 then B = -D --- (5)
(d2y/dx2)x=L= β2(AcoshβL + BsinhβL – CcosβL – DsinβL) = 0
Since β2 ≠ 0 and A = -C and B = -D and we have A(coshβL + cosβL) + B(sinhβL + sinβL) = 0 --- (6)
A/B =-(sinhβL + sinβL)/(coshβL + cosβL) ----- (7)
(d3y/dx3)x=L= β3(AsinhβL + BcoshβL + CsinβL - DcosβL) = 0
Again Since β2 ≠ 0 and A= -C and B= -D and we have A(sinhβL - sinβL) + B(coshβL + cosβL) = 0 ---- (8)
A/B =-(coshβL + cosβL)/(sinhβL - sinβL) ----- (9)
Dept. Of ME, ACE
12
Mechanical Vibrations [10ME72]
From equation (7) and (9) we have A(coshβL+cosβL) + B(sinhβL+sinβL) = 0 --- (10)
(sinhβL + sinβL) (coshβL + cosβL)
A/B = ---------------------- = ------------------------
(coshβL + cosβL) (sinhβL - sinβL)
Which reduces to coshβLcosβL + 1 = 0 ---------- (11)
Equation (11) is satisfied by a number of values of βL, corresponding to each normal mode of vibration,
which for the first three modes these values are tabulated in the table as
(β1l)2 = 3.52, (β2l)2 = 22.00, (β3l)2 = 61.70
Area A = 0.0012 m2 Moment of Inertia I = 1.600E-07m4
Density per unit length (ρ*A) = ρL = 7850x 0.0012 = 9.4200 Kg/m
Circular Frequency
ωn1 = (β1l)2√(EI)/(ρl4) = 2.052E+02 rad/sec
ωn2 = (β2l)2√(EI)/(ρl4) = 1.282E+03 rad/sec
ωn3= (β3l)2√(EI)/(ρl4) = 3.596E+03rad/sec
Natural Frequencies
fn1 = ωn1/2π = 32.639 Hz
fn2 = ωn2/2π = 203.994 Hz
fn3 = ωn3/2π = 572.111 Hz
Example-2: Determine the first three natural frequencies of a Rectangular Cantilever Beam for the
following data Length L = 1 m Breath B = 0.03 m Depth D = 0.04 m Young's Modulus E = 2.00E+11 N/m2
Density r = 7850.0 Kg/m3
Solution: For a cantilever beam the boundary conditions are at x= 0 displacement and slope are zero
Dept. Of ME, ACE 13
Mechanical Vibrations [10ME72]
i.e. y = 0 and dy/dx = 0 ------ (1)
and at x= L the shear force ‘V’ and the bending moment ‘M’ is zero
i.e. d2y/dx2 = 0 and d3y/dx3 = 0 -------- (2)
Substituting these boundary conditions into the general solution
Y=Acoshβx + Bsinhβx + Ccosβx + Dsinβx --- (3)
we obtain (y)x=0 = A + C = 0 then A = - C – (4)
(dy/dx)x=0= β(Asinhbx + Bcoshbx – Csinbx + Dcosbx)x=0 = 0
β(B + D) = 0 since β ≠ 0 then B = -D --- (5)
(d2y/dx2)x=L= β2(AcoshβL + BsinhβL – CcosβL – DsinβL) = 0
Since β2 ≠ 0 and A = -C and B = -D and we have
A(coshβL + cosβL) + B(sinhβL + sinβL) = 0 ---- (6)
A/B =-(sinhβL + sinβL)/(coshβL + cosβL) ----- (7)
(d3y/dx3)x=L= β3(AsinhβL + BcoshβL + CsinβL - DcosβL) = 0
Again Since β2 ≠ 0 and A = -C and B = -D and we have
A(sinhβL - sinβL) + B(coshβL + cosβL) = 0 ----- (8)
A/B =-(coshβL + cosβL)/(sinhβL - sinβL) ----- (9)
From equation (7) and (9) we have
A(coshβL+cosβL)+B(sinhβL+sinβL)= 0 --- (10)
(sinhβL + sinβL) (coshβL + cosβL)
A/B = ---------------------- = ------------------------
(coshβL + cosβL) (sinhβL - sinβL)
Which reduces to coshβLcosβL + 1 = 0 ---------- (11)
Dept. Of ME, ACE 14
Mechanical Vibrations [10ME72]
Equation (11) is satisfied by a number of values of βL, corresponding to each normal mode of vibration,
which for the first three modes these values are tabulated in the table as
(β1l)2 = 3.52, (β2l)2 = 22.00, (β3l)2 = 61.70
Area A = 0.0012 m2 Moment of Inertia I = 1.600E-07m4
Density per unit length (ρ*A) = ρL = 7850x 0.0012 = 9.4200 Kg/m
Circular Frequency
ωn1 = (β1l)2√(EI)/(ρl4) = 2.052E+02 rad/sec
ωn2 = (β2l)2√(EI)/(ρl4) = 1.282E+03 rad/sec
ωn3= (β3l)2√(EI)/(ρl4) = 3.596E+03rad/sec
Natural Frequencies
fn1 = ωn1/2π = 32.639 Hz
fn2 = ωn2/2π = 203.994 Hz
fn3 = ωn3/2π = 572.111 Hz
Dept. Of ME, ACE 15
